Year 4 kinematics: motion graphs

Mid-September. At the human-sciences high school (the Italian liceo delle scienze umane) and its economic-social option, Year 4 physics has just started again, and the first weeks are kinematics: uniform straight-line motion, equations of motion, distance-time and velocity-time graphs. With 2 hours of physics a week the programme does not linger: the teacher explains it in a few lessons and moves on to accelerated motion, then dynamics. The first test arrives in six to seven weeks, so between the end of October and the beginning of November.

There is a detail that makes this test different from what you expect: in maths, at the human-sciences high school, resolving into components with sine and cosine has not yet been covered. So the teacher cannot ask you for calculations with resolved vectors or angles — in one-dimensional kinematics they are not needed. What the test really asks is that you can read and produce graphs: understand that the slope of a straight line on an s-t graph is a velocity, and that the area under a broken line on a v-t graph is a displacement. That is where marks are lost, not in the formulas.

The three quantities that get confused

Before the formulas, three distinctions that the text of a problem often takes for granted, and which are instead the most common source of mistakes.

Displacement vs. distance travelled. Displacement is the difference between final and initial position, $Δ s = s_{f} - s_{i}$ : it has a sign, and if you return to the starting point it is zero even if you have walked for an hour. Distance travelled is the real length of the path, always positive, and takes no account of direction. A problem that asks “how far did it travel” and one that asks “what is its displacement” can have different answers even if they are about the same motion.

Average velocity vs. instantaneous velocity. Average velocity is defined over a time interval,

v_{m} = \frac{Δ s}{Δ t}

and it is a ratio between two overall numbers: it says nothing about what happens inside the interval. Instantaneous velocity is the value you would read on a speedometer at a precise instant $t$ . In uniform motion the two always coincide; in a motion that speeds up or slows down, they don’t — and this difference is exactly what produces the most surprising result in the last worked example below.

Acceleration. The most common mistake is to think “large acceleration” = “large velocity”. It is not so: acceleration is how much the velocity changes per unit of time,

a = \frac{Δ v}{Δ t}

A train travelling at a constant 200 km/h has zero acceleration. A bicycle that starts from rest and reaches 15 km/h in 3 seconds has an acceleration (small in absolute value, but not zero) that the train does not have.

Uniform motion and the s-t graph

In uniform straight-line motion the velocity is constant, and the equation of motion — the function that gives the position as a function of time — is:

s (t) = s_{0} + v t

where $s_{0}$ is the position at the instant $t = 0$ and $v$ is the velocity (constant, with the sign that indicates the direction). On the s-t graph this law is always a straight line, and its slope is exactly $v$ . Three readings that the test almost always asks for:

a line with positive slope → the moving object is moving away in the positive direction;
a horizontal line (zero slope) → the object is at rest, not “moving slowly”: zero slope means zero velocity, without exception;
a line with negative slope → the object is moving in the negative direction, that is, it is coming back towards the origin.

The problem of two objects meeting

It is the classic uniform-motion exercise that appears almost always in the first test: two objects with different equations of motion, asking when (and where) they meet comes down to setting the two equations against each other, that is, finding the instant at which $s_{A} (t) = s_{B} (t)$ . You can see it solved in full in the first exercise further down.

The v-t graph and uniformly accelerated motion

On the velocity-time graph, the information you need to read is different from that of the s-t graph: here it is the area under the curve that gives the displacement, not the slope. For a stretch at constant velocity the area is a rectangle; for an accelerated stretch, the shape is a trapezium (or a triangle, if you start from rest).

In uniformly accelerated motion the acceleration $a$ is constant, and on the v-t graph the velocity is therefore a straight line, with slope $a$ . The three laws you need are:

v (t) = v_{0} + a t

s (t) = s_{0} + v_{0} t + \frac{1}{2} a t^{2}

v^{2} = v_{0}^{2} + 2 a (s - s_{0})

The last one — the “time-free formula” — is the one you need every time the problem does not give (and does not ask for) the time taken, but only velocities and distance: it is the most frequent case in braking problems, which you see solved in the second exercise.

Free fall

Free fall is a particular case of uniformly accelerated motion with $a = g$ , where $g \approx 9.81 m/ s^{2}$ is the acceleration due to gravity. The only point where people stumble is the sign: if you choose the vertical axis positive upwards, a falling body has acceleration $- g$ ; if you choose it positive downwards, it has acceleration $+ g$ . There is no “right” convention in absolute terms — the right one is the one you choose at the start of the problem and keep until the end, without changing it halfway through the calculation. For a vertical throw upwards, the same time-free formula tells you the maximum height reached by a body (when $v = 0$ ) and the equation of motion tells you when it falls back to the ground.

The mistakes that cost marks

These are the mistakes I see coming back most often, with the wrong example and the correct one side by side.

1. Confusing the slope of s-t with the value of s. Mistake: reading an s-t graph and answering “at $t = 3 s$ the velocity is $12 m$ ” — a metre is not a velocity. Correct: the velocity is read as the ratio between the change in $s$ and the change in $t$ between two points of the line, that is, as the slope, and has the unit $m/s$ , not $m$ .

2. Using the uniform-motion formulas in an accelerated motion. Mistake: applying $s = v t$ to a body that is braking. If $v$ changes over time, $s = v t$ is not defined because there is no single $v$ . Correct: in uniformly accelerated motion you need the three laws of the previous section, not the one for uniform motion.

3. Forgetting the unit conversions and figures. Mistake: adding $54 km/h$ and $10 m/s$ without converting. Correct: first bring all quantities to the same unit. For km/h → m/s you divide by 3.6 ( $54 : 3.6 = 15 m/s$ ); for the opposite direction you multiply by 3.6.

4. The sign of $g$ . Mistake: using $+ g$ in a system with the positive axis upwards for a falling body, obtaining a height that grows instead of shrinking. Correct: fix the direction of the axis before writing any number, and use that sign for the whole exercise.

5. Reading the area under v-t as if it were a velocity. Mistake: answering “the displacement in the first 4 seconds is $10 m/s$ ” — that is a velocity value, not a distance. Correct: the area under a stretch of the v-t graph has the unit $m/s \cdot s = m$ , and it is a displacement, not a velocity.

Three worked examples

1. Two objects meeting in uniform motion

Two friends set off on bicycles towards each other from two villages $21 km$ apart. The first (A) leaves one of the two villages at $v_{A} = 30 km/h$ , the second (B) leaves the other village, in the opposite direction, at $v_{B} = 40 km/h$ . After how long do they meet, and at what distance from A’s village?

I fix the origin at A’s village, with the positive axis towards B. The two equations of motion are:

s_{A} (t) = 30 t s_{B} (t) = 21 - 40 t

The meeting happens when $s_{A} (t) = s_{B} (t)$ :

30 t = 21 - 40 t \Rightarrow 70 t = 21 \Rightarrow t = 0.3 h = 18 min

The position of the meeting, from A’s village:

s_{A} (0.3) = 30 \cdot 0.3 = 9 km

Check: from B, $s_{B} (0.3) = 21 - 40 \cdot 0.3 = 21 - 12 = 9 km$ — it matches, and it is the check that always works in these problems: the two equations of motion must give the same position at the instant of the meeting.

You will find more problems of this type, with the step-by-step correction, in the Year 2 kinematics test — it is called “Year 2” because at the science high school (liceo scientifico) kinematics is done in the second year, but the one-dimensional kinematics programme is the same: the correct answer is free, the guided solution unlocks with the first lesson.

2. Braking distance with the time-free formula

A car travels at $v_{0} = 90 km/h$ and brakes with constant deceleration $a = 5 m/ s^{2}$ until it stops. How far does it travel during braking?

First compulsory conversion: $90 km/h = 90 : 3.6 = 25 m/s$ . The problem does not give the braking time, so I use the time-free formula, with $v = 0$ (the car stops):

v^{2} = v_{0}^{2} - 2 a d \Rightarrow 0 = 2 5^{2} - 2 \cdot 5 \cdot d \Rightarrow d = \frac{625}{10} = 62.5 m

The minus sign in front of $2 a d$ is there because the car decelerates: the acceleration is opposite to the velocity.

3. Piecewise v-t graph: displacement and average velocity

A cyclist keeps $10 m/s$ for the first $2 s$ , then $4 m/s$ for the next $4 s$ (two horizontal stretches on the v-t graph, so two rectangles). How far does he travel in total, and does his average velocity coincide with the arithmetic mean of the two velocities?

The displacement is the sum of the two areas:

Δ s = 10 \cdot 2 + 4 \cdot 4 = 20 + 16 = 36 m

The total time is $6 s$ , so the average velocity is:

v_{m} = \frac{Δ s}{Δ t} = \frac{36}{6} = 6 m/s

The arithmetic mean of the two velocities would be $(10 + 4) /2 = 7 m/s$ : it does not coincide with $v_{m} = 6 m/s$ . The reason is that the two stretches last different times (2 s and 4 s): the average velocity weights each value by how long it lasted, the arithmetic mean does not. The two would coincide only if the two stretches had the same duration — it is not a special case to remember by heart, it is a direct consequence of the definition of average velocity as displacement divided by total time.

How we work on this topic

In my kinematics lessons I don’t start from the formula: I start from the graph. We draw a situation together (two moving objects, an acceleration, a braking), we build the equation of motion starting from the graph and, conversely, the graph starting from the equation of motion — because the test asks you for both directions, and being able to do only one of the two is the most common reason why an exercise “looks right” and then doesn’t add up. Before every calculation I also do a quick dimensional check: if the result of an area under v-t comes out in $m/s$ instead of metres, there is a mistake somewhere, and it is better to find it before handing in, not after. The method, more generally on how we work on physics online, is described on the physics page.

If you want to practise before the test, the Year 2 kinematics test covers exactly equations of motion and motion graphs. If you also need the picture of the most common mistakes in high-school physics in general, you will find the typical physics mistakes at high school, and if you are revising maths in parallel the Year 4 programme and formula sheet tells you what is coming in the next few months. Anyone in Year 3 who is tackling the resolution of forces without trigonometry yet will find the twin post in resolving forces and vectors without trigonometry.

If you want a hand before the October-November test, you can start with a free 30-minute introductory call: write to me.

Year 4 kinematics: equations of motion and graphs